Problem 22 - Calculus Early Transcendentals

Question

Use a line integral to find the area of the triangle with vertices \((0,0),(a, 0),\) and \((0, b),\) where \(a>0\) and \(b>0\)

Step-by-Step Solution

Answer

The area of the triangle is \(\frac{1}{2} ab\).

Step 1: Define the Problem

The problem requires finding the area of a triangle with vertices at \((0,0)\), \((a,0)\), and \((0,b)\) using a line integral. We will use Green's Theorem which relates a line integral around a simple closed curve to a double integral over the region it encloses.

Step 2: Set Up the Line Integral

By Green's Theorem, the area \(A\) enclosed by a path \(C\) can be found using the formula \[ A = \frac{1}{2} \oint_C (x \, dy - y \, dx). \] Here, \(C\) is the boundary of the triangle which runs along the path from \((0,0)\) to \((a,0)\), then to \((0,b)\), and back to \((0,0)\).

Step 3: Parameterize Each Edge of the Triangle

The path consists of three segments. For the first segment from \((0,0)\) to \((a,0)\), use \((x,y) = (t,0)\) for \(t\) from 0 to \(a\). For the second segment from \((a,0)\) to \((0,b)\), use \((x,y) = (a-t,b\frac{t}{a})\) for \(t\) from 0 to \(a\). For the third segment from \((0,b)\) back to \((0,0)\), use \((x,y) = (0,b-b\frac{t}{a})\) for \(t\) from 0 to \(a\).

Step 4: Calculate the Line Integral for Each Segment

For segment 1, \(x = t\), \(y = 0\), thus \(dy = 0\) and \(dx = dt\). The integral becomes \( \int_{0}^{a} (t (0) - 0) \, dt = 0\).For segment 2, \(x = a-t\), \(y = b \frac{t}{a}\), thus \(dy = \frac{b}{a} dt\) and \(dx = -dt\). The integral becomes \( \int_{0}^{a} ((a-t) \frac{b}{a} - b \frac{t}{a}) (-dt) = \int_{0}^{a} b dt = ba\).For segment 3, \(x = 0\), \(y = b-b\frac{t}{a}\), thus \(dy = -\frac{b}{a} dt\) and \(dx = 0\). The integral becomes \( \int_{0}^{a} (0 - (b-b\frac{t}{a})(0)) \, dt = 0\).

Step 5: Combine the Line Integral Results

Adding up the contributions from each segment, we find that the total integral is \(0 + ba + 0 = ba\). By Green's Theorem, the area is \( \frac{1}{2} ba \).

Step 6: Conclusion

The area of the triangle with vertices at \((0,0)\), \((a,0)\), and \((0,b)\) is \(\frac{1}{2} ab\) using the line integral method.

Key Concepts

Green's Theorem

Green's Theorem acts as a bridge between a line integral surrounding a closed curve and a double integral over the region that the curve encompasses. To apply this theorem to find the area of a shape, such as a triangle, you set up the line integral along the boundary of the shape. By using Green's Theorem, we transform a potentially complex line problem into a simpler double integral task over the region. In this specific problem, the theorem gives us this useful formula to find the area:\[ A = \frac{1}{2} \oint_C (x \, dy - y \, dx) \] Here, \(C\) refers to the path around the triangle's edges. You just need the right parameterization of each edge to perform the line integral accurately.

triangle area

The area of a triangle remains a fundamental concept in geometry, but here we explore it through the lens of calculus and Green's Theorem. Our triangle, with vertices at\((0,0)\), \((a,0)\), and \((0,b)\), can easily have its area calculated using standard geometry as \(\frac{1}{2}\) base \(\times\) height, which is \(\frac{1}{2}ab\). With Green's Theorem, we obtain the same result by applying the line integral approach. This method not only serves as a confirmation of the geometric area results but also deepens the understanding of the relationship between calculus and geometry. The closed path integration naturally includes subtractions and additions that ensure the calculation adheres strictly to the piecewise linear edges defining our triangle.

parametrization

Parametrization is a technique used to represent the equations of curves and paths in terms of a parameter, often denoted as \(t\). Each segment of the triangle's perimeter can be described as a unique pair of \(x(t)\) and \(y(t)\) equations. This transforms complex geometric paths into workable equations that link directly into computational methods such as line integration.For instance, during the parametrization:- **Segment 1**: Between \((0,0)\) and \((a,0)\), use \((x,y) = (t,0)\) where \(t\) ranges from \(0\) to \(a\).- **Segment 2**: From \((a,0)\) to \((0,b)\), we have \((x,y) = (a-t,b\frac{t}{a})\) with \(t\) in the same range.- **Segment 3**: Between \((0,b)\) back to \((0,0)\), use \((x,y) = (0,b-b\frac{t}{a})\).These parameterized equations lay the groundwork for evaluating the path integrals effectively, ensuring that the path followed during integration accurately traces the perimeter of the shape.